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1113002000183 is a prime number
BaseRepresentation
bin10000001100100100000…
…101101011111100110111
310221101212000121210221002
4100030210011223330313
5121213412003001213
62211150035313515
7143261140622423
oct20144405537467
93841760553832
101113002000183
1139a026682197
1215b85a14b89b
1380c566ca4c8
143bc261d3b83
151de4219a658
hex1032416bf37

1113002000183 has 2 divisors, whose sum is σ = 1113002000184. Its totient is φ = 1113002000182.

The previous prime is 1113002000119. The next prime is 1113002000189. The reversal of 1113002000183 is 3810002003111.

It is a strong prime.

It is an emirp because it is prime and its reverse (3810002003111) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1113002000183 - 26 = 1113002000119 is a prime.

It is a super-2 number, since 2×11130020001832 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1113002000189) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 556501000091 + 556501000092.

It is an arithmetic number, because the mean of its divisors is an integer number (556501000092).

Almost surely, 21113002000183 is an apocalyptic number.

1113002000183 is a deficient number, since it is larger than the sum of its proper divisors (1).

1113002000183 is an equidigital number, since it uses as much as digits as its factorization.

1113002000183 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 144, while the sum is 20.

Adding to 1113002000183 its reverse (3810002003111), we get a palindrome (4923004003294).

The spelling of 1113002000183 in words is "one trillion, one hundred thirteen billion, two million, one hundred eighty-three", and thus it is an aban number.